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Let $X_1, X_2, \ldots X_n$ identically distributed independent random variables with zero mean and finite variance $\sigma^2$. Let $\bar X$ be the sample mean and consider the random vector $ B= n^{-1/2}(X_1 - \bar X, \ldots , X_n - \bar X)$. Is it possible to calculate $$ E\left( \lambda_{max} (B^tB) \right) $$ Here $\lambda_{max}$ stands for the largest eigenvalue of a positive definite matrix.

It is straight forward to check that $\lambda_{max}( E(B^tB) ) = \frac{n-1}{n^2} \sigma^2$. But of course Jensen inequality works the other way around.

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Yes.

Because $rank(B^tB) = 1$ with probability one the only eigenvector with non null eigenvalue is $B^t$. In fact $ B^t B B^t = \|B\|^2 B^t $ , hence

$$ E ( \lambda_{max}(B^tB) ) = E( \|B\|^2 ) =\frac{1}{n} \sum_{i=1}^n E( X_i - \bar X )^2 = \frac{n-1}{n} \sigma^2 $$

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